Answer
$1.66\times10^{8}ms^{-1}$
Work Step by Step
We know: $\frac{n_{A}}{n_{B}}= 1.33$
$\frac{n_{A}}{n_{B}}= \frac{\frac{c}{v_{A}}}{\frac{c}{v_{B}}}= \frac{v_{B}}{v_{A}}$
Thus, we find:
1.33= $\frac{v_{B}}{1.25\times10^{8}m/s}$
Or $v_{B}= 1.33\times 1.25\times10^{8}m/s$ =$1.66\times10^{8}ms^{-1}$.
So, the speed of light in material B is $1.66\times10^{8}ms^{-1}$
